A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem.
Express your answer in terms of the variables M and a .

The side length will be a/4, and the center of a side will be dispalced a/8 from the center of the square. The moment of inertia, I, will be four times the value for one of the sides. For that value, you need to use the parallel axis theorem.

The moment of inertia of a single side rotated about its center of mass is
Icm = (1/12)*(mass of side)*(length of side)^2
= (1/12)(M/4)(a/4)^2 = (Ma^2/768

When rotated about the center of the square, you must add
(M/4)(a/8)^2 = Ma^2/256

Add those two together to get
Ma^2/192, and multiply by 4.

I get (1/48)M a^2

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